Question

Помогите пожалуйста даю 25 баллов

Answer 1

Ответ:

1. $-\frac{1}{x^{2} }$

2. $\frac{3y+3}{y}$

3. -a + 5.

Пошаговое объяснение:

1. $(\frac{x+4}{x-4} - \frac{x-4}{x+4} ) : \frac{16x^{3} }{16-x^{2} }$ = $\frac{(x+4)^2-(x-4)^2}{(x-4)*(x+4)} * \frac{16-x^2}{16x^3}$ = $\frac{8*2x}{(x-4)(x+4)} * \frac{(4-x)(4+x)}{16x^3}$ = $\frac{16x}{x-4} * \frac{-(x-4)*1}{16x^3}$ = $\frac{-1}{x^2}$ .

2. $\frac{3y}{y-1} - \frac{y+3}{5y-5} * \frac{15}{y^2+3y}$ = $\frac{3y}{y-1} - \frac{y+3}{5(y-1)} * \frac{15}{y*(y+3)}$ = $\frac{3y}{y-1} - \frac{1}{y-1} * \frac{3}{y}$ = $\frac{3y}{y-1} - \frac{3}{y*(y-1)}$ = $\frac{3y^2-3}{y*(y-1)}$ = $\frac{3(y^2-1)}{y*(y-1)}$ = $\frac{3(y-1)(y+1)}{y(y-1)}$ = $\frac{3(y+1)}{y}$ = $\frac{3y+3}{y}$ .

3. $\frac{a-\frac{10a-25}{a} }{\frac{5}{a}-1 }$ = $\frac{\frac{a^2-(10a-25)}{a} }{\frac{5-a}{a} }$ = $\frac{a^2-10a+25}{5-a}$ = $\frac{(a-5)^2}{-(a-5)}$ = -(a - 5) = -a + 5.